TITLE: When does Pontryagin duality generalize?
QUESTION [11 upvotes]: Let $T$ be a locally compact abelian (LCA) group. For any other LCA group $G$, let 
$\hom(G,T)$ be the set of continuous homomorphisms $G\to T$. With the compact-open 
topology, $\hom(G,T)$ is certainly a topological group, but is not in general locally 
compact, even if $T$ is compact. In any case there is an obvious homomorphism 
$$
  \alpha_G : G \to \hom(\hom(G,T),T) 
$$
sending $x\in G$ to the functional $\chi\mapsto \chi(x)$. 
I have three questions: 

For what LCA groups $T$ is $\hom(G,T)$ locally compact for all locally compact 
      $G$?
For what groups $T$ is $\alpha_G$ always a (topological) isomorphism?
  More generally, is there any full subcategory $\mathcal L$ of the category of LCA 
      groups for which $\hom(G,H)$ is in $\mathcal L$ whenever $G$ and $H$ are in 
      $\mathcal L$?

(Edit: I'm most interested in 1 and 2. If there is some $\mathcal L$ satisfying 3 
where the objects in $\mathcal L$ can be characterized by some interesting topological 
criterion, that would be fantastic. My question is not whether there are "ad-hoc" ways of constructing $\mathcal L$.)

REPLY [12 votes]: I gave an answer to the second question already on math.stackexchange at https://math.stackexchange.com/questions/124379/why-unitary-characters-for-the-dual-group-in-pontryagin-duality-if-g-is-not-co, but I'll put it here too: if $\alpha_G$ is an isomorphism of topological groups for all locally compact abelian groups $G$ then $T$ must be the unit circle. This explains the importance of the unit circle as a target group. It is discussed in Pontryagin's book on topological groups (see Example 72) and the first volume on abstract harmonic analysis by Hewitt and Ross (p. 424).

REPLY [9 votes]: Here's an answer for 1: given a LCA group $T$, $\text{Hom}(G,T)$ is LC for every LCA $G$ iff $T$ is a compact Lie group.
Proof: Clearly this is a sufficient condition, because it holds for connected tori, and is obviously stable under taking closed subgroup.
Conversely suppose the condition is satisfied.
Let $\mathbf{Z}^{(\mathbf{N})}$ denote the discrete free abelian group of countable rank. Then $\text{Hom}(\mathbf{Z}^{(\mathbf{N})},T)=T^{\mathbf{N}}$ is locally compact iff $T$ is compact. 
So $T$ is compact. Let now $D$ be the Pontryagin dual of $T$, so that $D$ is discrete. By the condition and Pontryagin duality, $\text{Hom}(D,G)$ is LC for all LCA $G$.
Consider a discrete abelian group $G_0$ (I'll fix it later). For $F\subset D$ finite subset, define $U_F$ as the set of homomorphisms $D\to G_0$ vanishing on $F$. So the $U_F$ form a basis of clopen neighbourhoods of 0 in $\text{Hom}(D,G_0)$. So one of those, say $U_F$ is compact. If $E$ is the subgroup generated by $F$, this shows that $\text{Hom}(D/E,G_0)$ is compact. Now to specify, let us assume from the beginning we picked the group $G_0=H^{(\mathbf{N})}$, where $H=\mathbf{Q}/\mathbf{Z}$. It is a standard verification that if $D/E$ is nonzero, then $\text{Hom}(D/E,H)$ is not trivial, and an easy consequence is that $\text{Hom}(D/E,G_0)$ is not compact, a contradiction. This shows that $D/E=0$. So $D$ is finitely generated, and thus $T$ is a compact abelian Lie group, as required.
A careful look on the proof shows that otherwise, $\text{Hom}(G,T)$ is not locally compact for the specific choice of $G=\hat{\mathbf{Z}}^{\mathbf{N}}\times\mathbf{Z}^{(\mathbf{N})}$, where $\hat{\mathbf{Z}}=\prod_p\mathbf{Z}_p$ and $\mathbf{Z}_p$ are the $p$-adics.
Edit: the answer for 2. is, as mentioned in another post, that only the circle satisfies the condition. (I initially thought it was trivially true for the 2-torus (or even the trivial group!), while it's trivially false.)
Here's a proof. Taking $G=\mathbf{R}/\mathbf{Z}$, the condition implies that $\text{Hom}(\mathbf{R}/\mathbf{Z},T)$ is nonzero; pick $f$; its kernel is finite and the quotient of $\mathbf{R}/\mathbf{Z}$ by the kernel of $f$ is also isomorphic to the circle, so the image is isomorphic to the circle. Now by Pontryagin duality, whenever $\mathbf{R}/\mathbf{Z}$ stands as a closed subgroup in a LCA group, it stands as a direct factor. So write $T=H\times \mathbf{R}/\mathbf{Z}$. Then for every $G$ we get $\text{Hom}(\text{Hom}(G,T),T)=G\times \text{Hom}(\hat{G},H)\times \widehat{\text{Hom}(G,H)}\times\text{Hom}(\text{Hom}(G,H),H)$. Since by assumption the identity of $G$ into the first factor is an isomorphism for every $G$, it follows that all other 3 factors are zero. Taking $G=\mathbf{Z}$, we deduce $\widehat{\text{Hom}(\mathbf{Z},H)}=0$, so $H=\text{Hom}(\mathbf{Z},H)=0$ by Pontryagin duality, hence $T=\mathbf{R}/\mathbf{Z}$.

REPLY [5 votes]: Piggy-backing on the excellent answer by Yves, I think $\mathbb{R}/\mathbb{Z}$ is the only answer for $T$ in question 2. 
Every compact Lie group $T$ is isomorphic to one of the form $F \times (\mathbb{R}/\mathbb{Z})^n$ for a finite abelian group $F$ and finite $n \geq 0$. A necessary condition for satisfaction of 2. is that the canonical map $\mathbb{Z} \to \hom(T, T)$ is an isomorphism. 
We then have 
$$\hom(T, T) = \hom(T, F \times (\mathbb{R}/\mathbb{Z})^n) \cong \hom(T, \mathbb{R}/\mathbb{Z})^n \times \hom(T, F)$$ 
which contains $\hom((\mathbb{R}/\mathbb{Z})^n, \mathbb{R}/\mathbb{Z})^n \cong \mathbb{Z}^{n^2}$ as an infinite summand (with a complementary summand that is finite). Thus $n = 1$. Next, if $F$ is non-trivial, we have a product decomposition 
$$\hom(T, T) = \hom(\mathbb{R}/\mathbb{Z}, \mathbb{R}/\mathbb{Z}) \times \hom(\mathbb{R}/\mathbb{Z}, F) \times \hom(F, \mathbb{R}/\mathbb{Z}) \times \hom(F, F)$$ 
for which the third and fourth factors are non-trivial, and hence $\mathbb{Z}$ cannot project onto the displayed product. Thus, $F = 0$, which completes the proof. 
Edit: I might as well promote my earlier comment as an answer to question 3, as efficiently reformulated by Yves Cornulier: a suitable full subcategory of LCA-groups that is closed under homs is the category of compactly generated abelian Lie groups. (I make no claim is that this is maximal among subcategories that are hom-closed, however.) 
Edit 2: Actually, looking back on an old $n$-Category Café discussion on related matters, John Baez mentions a paper by Michael Barr which in some sense blows my answer to 3 out of the water, and which I'm sure Daniel Miller would be interested in. Here is the abstract: 

Abstract. Let $\mathcal{G}$ denote the full subcategory of topological abelian groups consisting of the groups that can be embedded algebraically and topologically into a product of locally compact abelian groups. We show that there is a full coreflective subcategory $\mathcal{S}$ of $\mathcal{G}$ that contains all locally compact groups and is $\ast$-autonomous. This means that for all $G$, $H$ in $\mathcal{S}$ there is an “internal hom” $G \multimap H$ whose underlying abelian group is $\hom(G, H)$ and that that makes $\mathcal{S}$ into a closed category with a tensor product whose underlying abelian group is a quotient of the algebraic tensor product. Moreover a perfect duality results if we let $T$ denote the circle group and define $G^\ast = G \multimap T$. This is essentially a new exposition of work originally done jointly with H. Kleisli [Theory Appl. Categories, 8, 54–62].